Let's focus on the case where all the closed fibers are $\mathbf{P}^1$. I'm not sure if it's automatic, so I will also assume the generic fibre to be $\mathbf{P}^1$. Then, by flatness, all the fibres are (possible singular) curves over $\mathbf{C}$. The arithmetic genus is constant in the fibres. Also, the morphism is smooth over the set of closed points. Therefore it is smooth everywhere. In particular, all the fibres are $\mathbf{P}^1$.
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Ariyan JavanpeykarNov 30 '11 at 10:52

Thank you Ariyan. I don't understand why you say that all the fibers in this case are schemes over $\mathbb{C}$.
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origalNov 30 '11 at 11:29

Hi origal. I don't understand it either. Please ignore that. It should read, the fibre over $y$ in $Y$ is a (possibly singular) curve over the residue field $k(y)$. (For some reason, this morning, I thought this would always be $\mathbf{C}$.) The conclusion then reads that the fibre over $y$ is $\mathbf{P}^1_{k(y)}$. In view of Laurent Moret-Bailly's answer below, note that we do not use that $Y$ is normal.
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Ariyan JavanpeykarNov 30 '11 at 13:30

I think there is one more problem: what is constant in a flat family is the arithmetic genus, then, in this case $h^0(X_y,\mathcal{O}_{X_y})-h^1(X_y,\mathcal{O}_{X_y})=1$ (with $X_y$ fiber over $y$). But is it true that $h^0(X_y,\mathcal{O}_{X_y})=1$ if the variety $X_y$ is not defined over $\mathbb{C}$? I think we need $k(y)$ to be algebraically closed.
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origalNov 30 '11 at 15:17

1 Answer
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The fibers are normal. This follows (without the normality assumption on $Y$) from EGA IV (12.1.6) which says that the set those $x\in X$ where the fiber is normal is open, hence if its complement were nonempty, it would contain a closed point.